Question

7 Reaction Yields OLD Ас Np Pu Am Cm Bk Cf Es if 11.7 g of aluminum reacts with 37.2 g of copper (II) sulfate according to the following reaction, how many grams of aluminum sulfate will be produced? 2A1 + 3CuSO4 + Al(SO4)3 + 3Cu Your answer should have three significant figures. (Round your answer to the first decimal place.) Provide your answer below:

Answer 1

Number of moles of Al = 11.7 g / 26.9815 g/mol = 0.434 mole

Number of moles of CuSO4 = 37.2 g / 159.609 g/mol = 0.233 mole

From the balanced equation we can say that

2 mole of Al requires 3 mole of CuSO4 so

0.434 mole of Al will require

= 0.434 mole of Al *(3 mole of CuSO4 / 2 mole of Al)

= 0.651 mole of CuSO4

But we have 0.233 mole of CuSO4 which is in short so CuSO4 is limiting reactant

From the balanced equation we can say that

3 mole of CuSO4 produces 1 mole of Al2(SO4)3 so

0.233 mole of CuSO4 will produce

= 0.233 mole of CuSO4 *(1 mole of Al2(SO4)3 / 3 mole of CuSO4)

= 0.0777 mole of Al2(SO4)3

mass of 1 mole of Al2(SO4)3 = 342.15 g so

the mass of 0.0777 mole of Al2(SO4)3 = 26.6 g

Therefore, the mass of Al2(SO4)3 produced would be 26.6 g